Difference between revisions of "2024 AMC 10A Problems/Problem 22"
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Let <math>\mathcal K</math> be the kite formed by joining two right triangles with legs <math>1</math> and <math>\sqrt3</math> along a common hypotenuse. Eight copies of <math>\mathcal K</math> are used to form the polygon shown below. What is the area of triangle <math>\Delta ABC</math>? | Let <math>\mathcal K</math> be the kite formed by joining two right triangles with legs <math>1</math> and <math>\sqrt3</math> along a common hypotenuse. Eight copies of <math>\mathcal K</math> are used to form the polygon shown below. What is the area of triangle <math>\Delta ABC</math>? | ||
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<math>\textbf{(A) }2+3\sqrt3\qquad\textbf{(B) }\dfrac92\sqrt3\qquad\textbf{(C) }\dfrac{10+8\sqrt3}{3}\qquad\textbf{(D) }8\qquad\textbf{(E) }5\sqrt3</math> | <math>\textbf{(A) }2+3\sqrt3\qquad\textbf{(B) }\dfrac92\sqrt3\qquad\textbf{(C) }\dfrac{10+8\sqrt3}{3}\qquad\textbf{(D) }8\qquad\textbf{(E) }5\sqrt3</math> | ||
Revision as of 17:38, 8 November 2024
Problem
Let 
be the kite formed by joining two right triangles with legs 
and 
along a common hypotenuse. Eight copies of are used to form the polygon shown below. What is the area of triangle 
?
Solution
Let be quadrilateral MNOP. Drawing line MO splits the triangle into 
.
Drawing the altitude from N to point Q on line MO, we know NQ is 
, MQ is 
, and QO is 
.
See also
| 2024 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
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